# Question #0c86a

Jun 21, 2016

Discussed below

#### Explanation:

Given

$h \to \text{The height from which the canon ball was fired} = 60 m$

$u \to \text{The vlocity of projection} = 25 \frac{m}{s}$

$\alpha \to \text{Angle of projection} = {53}^{\circ}$

$\text{And } \sin {53}^{\circ} = 0.8 \mathmr{and} \cos {53}^{\circ} = 0.6$

Now

$\text{Horizontal component of vel.of projection } u \cos {53}^{\circ} = 25 \cdot 0.6 = 15 \frac{m}{s}$

$\text{Vertical component of vel.of projection } u \sin {53}^{\circ} = 25 \cdot 0.8 = 20 \frac{m}{s}$

At the time when it reaches at maximum height H from its point of projection its vertical component becomes zero, So considering $g = m {s}^{-} 2$ we get

${0}^{2} = {20}^{2} - 2 \cdot 10 \cdot H \implies H = 20 m$

Hence at its maximum height the total height fron ground
${H}_{\text{total}} = 60 + 20 = 80 m$

Now if its time of flight be t s
then it will undergo a vertical displacement downward during this time. So
-
$- 60 = 20 \cdot t - \frac{1}{2} \cdot 20 \cdot {t}^{2}$

$\implies {t}^{2} - 2 t - 6 = 0$

$\implies t = \frac{2 + \sqrt{4 - 4 \cdot \left(- 6\right) \cdot 1}}{2}$

$\implies t = \left(\sqrt{7} + 1\right) s$

Net horizontal displacement during this time of flight

$= u \cos \alpha \cdot t = 15 \left(\sqrt{7} + 1\right) m$